\(\int \frac {e^{\frac {45+(25 x^2+10 x^3+x^4) \log (\frac {19+x}{3})}{x^2 \log (\frac {19+x}{3})}} (-45 x+(-1710-90 x) \log (\frac {19+x}{3})+(190 x^3+48 x^4+2 x^5) \log ^2(\frac {19+x}{3}))}{(19 x^3+x^4) \log ^2(\frac {19+x}{3})} \, dx\) [6603]

Optimal result

Integrand size = 108, antiderivative size = 25 \[ \int \frac {e^{\frac {45+\left (25 x^2+10 x^3+x^4\right ) \log \left (\frac {19+x}{3}\right )}{x^2 \log \left (\frac {19+x}{3}\right )}} \left (-45 x+(-1710-90 x) \log \left (\frac {19+x}{3}\right )+\left (190 x^3+48 x^4+2 x^5\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{\left (19 x^3+x^4\right ) \log ^2\left (\frac {19+x}{3}\right )} \, dx=e^{(5+x)^2+\frac {45}{x^2 \log \left (4+\frac {7+x}{3}\right )}} \]

[Out]

exp((5+x)^2+45/x^2/ln(1/3*x+19/3))

Rubi [A] (verified)

Time = 3.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {1607, 6873, 6820, 6838} \[ \int \frac {e^{\frac {45+\left (25 x^2+10 x^3+x^4\right ) \log \left (\frac {19+x}{3}\right )}{x^2 \log \left (\frac {19+x}{3}\right )}} \left (-45 x+(-1710-90 x) \log \left (\frac {19+x}{3}\right )+\left (190 x^3+48 x^4+2 x^5\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{\left (19 x^3+x^4\right ) \log ^2\left (\frac {19+x}{3}\right )} \, dx=e^{\frac {45}{x^2 \log \left (\frac {x+19}{3}\right )}+(x+5)^2} \]

[In]

Int[(E^((45 + (25*x^2 + 10*x^3 + x^4)*Log[(19 + x)/3])/(x^2*Log[(19 + x)/3]))*(-45*x + (-1710 - 90*x)*Log[(19
+ x)/3] + (190*x^3 + 48*x^4 + 2*x^5)*Log[(19 + x)/3]^2))/((19*x^3 + x^4)*Log[(19 + x)/3]^2),x]

[Out]

E^((5 + x)^2 + 45/(x^2*Log[(19 + x)/3]))

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {45+\left (25 x^2+10 x^3+x^4\right ) \log \left (\frac {19+x}{3}\right )}{x^2 \log \left (\frac {19+x}{3}\right )}\right ) \left (-45 x+(-1710-90 x) \log \left (\frac {19+x}{3}\right )+\left (190 x^3+48 x^4+2 x^5\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{x^3 (19+x) \log ^2\left (\frac {19+x}{3}\right )} \, dx \\ & = \int \frac {\exp \left (\frac {45+\left (25 x^2+10 x^3+x^4\right ) \log \left (\frac {19+x}{3}\right )}{x^2 \log \left (\frac {19}{3}+\frac {x}{3}\right )}\right ) \left (-45 x+(-1710-90 x) \log \left (\frac {19+x}{3}\right )+\left (190 x^3+48 x^4+2 x^5\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{x^3 (19+x) \log ^2\left (\frac {19}{3}+\frac {x}{3}\right )} \, dx \\ & = \int \frac {e^{(5+x)^2+\frac {45}{x^2 \log \left (\frac {19+x}{3}\right )}} \left (-45 x-90 (19+x) \log \left (\frac {19+x}{3}\right )+2 x^3 \left (95+24 x+x^2\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{x^3 (19+x) \log ^2\left (\frac {19}{3}+\frac {x}{3}\right )} \, dx \\ & = e^{(5+x)^2+\frac {45}{x^2 \log \left (\frac {19+x}{3}\right )}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {45+\left (25 x^2+10 x^3+x^4\right ) \log \left (\frac {19+x}{3}\right )}{x^2 \log \left (\frac {19+x}{3}\right )}} \left (-45 x+(-1710-90 x) \log \left (\frac {19+x}{3}\right )+\left (190 x^3+48 x^4+2 x^5\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{\left (19 x^3+x^4\right ) \log ^2\left (\frac {19+x}{3}\right )} \, dx=e^{25+10 x+x^2+\frac {45}{x^2 \log \left (\frac {19+x}{3}\right )}} \]

[In]

Integrate[(E^((45 + (25*x^2 + 10*x^3 + x^4)*Log[(19 + x)/3])/(x^2*Log[(19 + x)/3]))*(-45*x + (-1710 - 90*x)*Lo
g[(19 + x)/3] + (190*x^3 + 48*x^4 + 2*x^5)*Log[(19 + x)/3]^2))/((19*x^3 + x^4)*Log[(19 + x)/3]^2),x]

[Out]

E^(25 + 10*x + x^2 + 45/(x^2*Log[(19 + x)/3]))

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\left (x^{4}+10 x^{3}+25 x^{2}\right ) \ln \left (\frac {x}{3}+\frac {19}{3}\right )+45}{x^{2} \ln \left (\frac {x}{3}+\frac {19}{3}\right )}}\)	\(37\)
risch	\({\mathrm e}^{\frac {\ln \left (\frac {x}{3}+\frac {19}{3}\right ) x^{4}+10 \ln \left (\frac {x}{3}+\frac {19}{3}\right ) x^{3}+25 x^{2} \ln \left (\frac {x}{3}+\frac {19}{3}\right )+45}{x^{2} \ln \left (\frac {x}{3}+\frac {19}{3}\right )}}\)	\(48\)

[In]

int(((2*x^5+48*x^4+190*x^3)*ln(1/3*x+19/3)^2+(-90*x-1710)*ln(1/3*x+19/3)-45*x)*exp(((x^4+10*x^3+25*x^2)*ln(1/3
*x+19/3)+45)/x^2/ln(1/3*x+19/3))/(x^4+19*x^3)/ln(1/3*x+19/3)^2,x,method=_RETURNVERBOSE)

[Out]

exp(((x^4+10*x^3+25*x^2)*ln(1/3*x+19/3)+45)/x^2/ln(1/3*x+19/3))

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {45+\left (25 x^2+10 x^3+x^4\right ) \log \left (\frac {19+x}{3}\right )}{x^2 \log \left (\frac {19+x}{3}\right )}} \left (-45 x+(-1710-90 x) \log \left (\frac {19+x}{3}\right )+\left (190 x^3+48 x^4+2 x^5\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{\left (19 x^3+x^4\right ) \log ^2\left (\frac {19+x}{3}\right )} \, dx=e^{\left (\frac {{\left (x^{4} + 10 \, x^{3} + 25 \, x^{2}\right )} \log \left (\frac {1}{3} \, x + \frac {19}{3}\right ) + 45}{x^{2} \log \left (\frac {1}{3} \, x + \frac {19}{3}\right )}\right )} \]

[In]

integrate(((2*x^5+48*x^4+190*x^3)*log(1/3*x+19/3)^2+(-90*x-1710)*log(1/3*x+19/3)-45*x)*exp(((x^4+10*x^3+25*x^2
)*log(1/3*x+19/3)+45)/x^2/log(1/3*x+19/3))/(x^4+19*x^3)/log(1/3*x+19/3)^2,x, algorithm="fricas")

[Out]

e^(((x^4 + 10*x^3 + 25*x^2)*log(1/3*x + 19/3) + 45)/(x^2*log(1/3*x + 19/3)))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {45+\left (25 x^2+10 x^3+x^4\right ) \log \left (\frac {19+x}{3}\right )}{x^2 \log \left (\frac {19+x}{3}\right )}} \left (-45 x+(-1710-90 x) \log \left (\frac {19+x}{3}\right )+\left (190 x^3+48 x^4+2 x^5\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{\left (19 x^3+x^4\right ) \log ^2\left (\frac {19+x}{3}\right )} \, dx=e^{\frac {\left (x^{4} + 10 x^{3} + 25 x^{2}\right ) \log {\left (\frac {x}{3} + \frac {19}{3} \right )} + 45}{x^{2} \log {\left (\frac {x}{3} + \frac {19}{3} \right )}}} \]

[In]

integrate(((2*x**5+48*x**4+190*x**3)*ln(1/3*x+19/3)**2+(-90*x-1710)*ln(1/3*x+19/3)-45*x)*exp(((x**4+10*x**3+25
*x**2)*ln(1/3*x+19/3)+45)/x**2/ln(1/3*x+19/3))/(x**4+19*x**3)/ln(1/3*x+19/3)**2,x)

[Out]

exp(((x**4 + 10*x**3 + 25*x**2)*log(x/3 + 19/3) + 45)/(x**2*log(x/3 + 19/3)))

Maxima [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {45+\left (25 x^2+10 x^3+x^4\right ) \log \left (\frac {19+x}{3}\right )}{x^2 \log \left (\frac {19+x}{3}\right )}} \left (-45 x+(-1710-90 x) \log \left (\frac {19+x}{3}\right )+\left (190 x^3+48 x^4+2 x^5\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{\left (19 x^3+x^4\right ) \log ^2\left (\frac {19+x}{3}\right )} \, dx=e^{\left (x^{2} + 10 \, x - \frac {45}{x^{2} {\left (\log \left (3\right ) - \log \left (x + 19\right )\right )}} + 25\right )} \]

[In]

integrate(((2*x^5+48*x^4+190*x^3)*log(1/3*x+19/3)^2+(-90*x-1710)*log(1/3*x+19/3)-45*x)*exp(((x^4+10*x^3+25*x^2
)*log(1/3*x+19/3)+45)/x^2/log(1/3*x+19/3))/(x^4+19*x^3)/log(1/3*x+19/3)^2,x, algorithm="maxima")

[Out]

e^(x^2 + 10*x - 45/(x^2*(log(3) - log(x + 19))) + 25)

Giac [A] (verification not implemented)

none

Time = 0.65 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {45+\left (25 x^2+10 x^3+x^4\right ) \log \left (\frac {19+x}{3}\right )}{x^2 \log \left (\frac {19+x}{3}\right )}} \left (-45 x+(-1710-90 x) \log \left (\frac {19+x}{3}\right )+\left (190 x^3+48 x^4+2 x^5\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{\left (19 x^3+x^4\right ) \log ^2\left (\frac {19+x}{3}\right )} \, dx=e^{\left (x^{2} + 10 \, x + \frac {45}{x^{2} \log \left (\frac {1}{3} \, x + \frac {19}{3}\right )} + 25\right )} \]

[In]

integrate(((2*x^5+48*x^4+190*x^3)*log(1/3*x+19/3)^2+(-90*x-1710)*log(1/3*x+19/3)-45*x)*exp(((x^4+10*x^3+25*x^2
)*log(1/3*x+19/3)+45)/x^2/log(1/3*x+19/3))/(x^4+19*x^3)/log(1/3*x+19/3)^2,x, algorithm="giac")

[Out]

e^(x^2 + 10*x + 45/(x^2*log(1/3*x + 19/3)) + 25)

Mupad [B] (verification not implemented)

Time = 11.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {45+\left (25 x^2+10 x^3+x^4\right ) \log \left (\frac {19+x}{3}\right )}{x^2 \log \left (\frac {19+x}{3}\right )}} \left (-45 x+(-1710-90 x) \log \left (\frac {19+x}{3}\right )+\left (190 x^3+48 x^4+2 x^5\right ) \log ^2\left (\frac {19+x}{3}\right )\right )}{\left (19 x^3+x^4\right ) \log ^2\left (\frac {19+x}{3}\right )} \, dx={\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{25}\,{\mathrm {e}}^{\frac {45}{x^2\,\ln \left (\frac {x}{3}+\frac {19}{3}\right )}} \]

[In]

int(-(exp((log(x/3 + 19/3)*(25*x^2 + 10*x^3 + x^4) + 45)/(x^2*log(x/3 + 19/3)))*(45*x + log(x/3 + 19/3)*(90*x
+ 1710) - log(x/3 + 19/3)^2*(190*x^3 + 48*x^4 + 2*x^5)))/(log(x/3 + 19/3)^2*(19*x^3 + x^4)),x)

[Out]

exp(10*x)*exp(x^2)*exp(25)*exp(45/(x^2*log(x/3 + 19/3)))